AbigailII: Ok, I kind of meant that without being clear. so i see further down in the thread you describe an algorithm that works for the simple case.
So if T=W+D/2 for the leader, and T(n)=W+D/2+U for the others for simplicity

So if you are searching to see if T > T(n) for all n

If it doesn't then you need to see if T=Tn for each value of n. If it does then you would just search through a tree to determine if any combination of results ends up with the leaders S-B score being less than the challengers. That's what I meant by being efficient. You don't need to take into account every possible outcome.